The ionization enthalpy of hydrogen atom is `1.312 xx 10^6 J mol^-1`. The energy required to excite the electron in the atom from `n= 1` to `n = 2` is :

Options

A. `9.84xx10^(5) J "mol"^(-1)`
B. `8.51xx10^(5) J "mol"^(-1)`
C. `6.56xx10^(5) J "mol"^(-1)`
D. `7.56xx10^(5) J "mol"^(-1)`

Answer

A

Explanation

`IE=E_(oo)-E_(1)`
`E_(1)=zero-IE_(1)=-1.312xx10^(6)"J mol"^(-1)`.
`E_(2)=(-1.312xx10^(6))/((2)^(2))=(-1.312xx10^(6))/(4)`
`=-1.312xx10^(6)`
`DeltaE=E_(2)-E_(1)=-1.312xx10^(6)((1)/(4)-1)`
`=(-1.312)xx(-0.75)xx10^(6)`
`=9.84xx10^(5)"J mol"^(-1)`.